Derivatives of Tangent and Reciprocal Trig Functions (College Board AP® Calculus BC): Revision Note

Derivative of the tangent function

What is the derivative of tan x?

• If $f\left(x\right)=\tan x$ then $f^{\text{'}}\left(x\right)={\sec }^{2}x$

• This can be shown using the identity $\tan x\equiv \frac{\sin x}{\cos x}$ and the quotient rule

  • The quotient rule states that if $y=\frac{u}{v}$, then $y^{\text{'}}=\frac{u^{\text{'}}v-u{v}^{\text{'}}}{v^2}$

  • Let $u=\sin x$ and $v=\cos x$

  • So $u^{\text{'}}=\cos x$ and $v^{\text{'}}=-\sin x$

  • Applying the quotient rule

    • $y^{\text{'}}=\frac{\cos x·\cos x-\sin x·\left(-\sin x\right)}{{\left(\cos x\right)}^2}$

  • Simplifying

    • $y^{\text{'}}=\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}$

  • Using the identity ${\sin }^{2}x+{\cos }^{2}x\equiv 1$

    • $y^{\text{'}}=\frac{1}{{\cos }^{2}x}$

  • This is the definition of ${\sec }^{2}x$, as $\sec x=\frac{1}{\cos x}$

    • $y^{\text{'}}={\sec }^{2}x$

What is the derivative of tan kx?

• If $f\left(x\right)=\tan kx$ then $f^{\text{'}}\left(x\right)=k{\sec }^{2}kx$

  • This is a result of applying the chain rule

Derivatives of reciprocal trig functions

What are the reciprocal trig functions?

• The reciprocal trigonometric functions are:

  • $\frac{1}{\sin x}=\csc x$

  • $\frac{1}{\cos x}=\sec x$

  • $\frac{1}{\tan x}=\cot x$

What are the derivatives of the reciprocal trig functions?

• If $f\left(x\right)=\csc x$ then $f^{\text{'}}\left(x\right)=-\cot x\csc x$

• If $g\left(x\right)=\sec x$ then $g^{\text{'}}\left(x\right)=\tan x\sec x$

• If $h\left(x\right)=\cot x$ then $h^{\text{'}}\left(x\right)=-{\csc }^{2}x$

• These results can be remembered, or they can be derived using the reciprocal trig function definitions and the quotient rule

• The table below shows the derivatives when $x$ is replaced with $kx$

How do I derive the derivative of csc x?

• Recall that $\csc x=\frac{1}{\sin x}$

• Apply the quotient rule, $y^{\text{'}}=\frac{u^{\text{'}}v-u{v}^{\text{'}}}{v^2}$

  • $u=1$ and $v=\sin x$

  • $u^{\text{'}}=0$ and $v^{\text{'}}=\cos x$

  • $y^{\text{'}}=\frac{0-\cos x}{{\left(\sin x\right)}^2}$

• Simplify using the identities $\frac{\cos x}{\sin x}=\cot x$ and $\frac{1}{\sin x}=\csc x$

  • $y^{\text{'}}=\frac{-\cos x}{{\sin }^{2}x}=-\frac{\cos x}{\sin x}·\frac{1}{\sin x}=-\cot x\csc x$

How do I derive the derivative of sec x?

• Recall that $\sec x=\frac{1}{\cos x}$

• Apply the quotient rule, $y^{\text{'}}=\frac{u^{\text{'}}v-u{v}^{\text{'}}}{v^2}$

  • $u=1$ and $v=\cos x$

  • $u^{\text{'}}=0$ and $v^{\text{'}}=-\sin x$

  • $y^{\text{'}}=\frac{0-\left(-\sin x\right)}{{\left(\cos x\right)}^2}$

• Simplify using the identities $\frac{\sin x}{\cos x}=\tan x$ and $\frac{1}{\cos x}=\sec x$

  • $y^{\text{'}}=\frac{\sin x}{{\cos }^{2}x}=\frac{\sin x}{\cos x}·\frac{1}{\cos x}=\tan x\sec x$

How do I derive the derivative of cot x?

• Recall that $\cot x=\frac{1}{\tan x}=\frac{\cos x}{\sin x}$

• Apply the quotient rule, $y^{\text{'}}=\frac{u^{\text{'}}v-u{v}^{\text{'}}}{v^2}$

  • $u=\cos x$ and $v=\sin x$

  • $u^{\text{'}}=-\sin x$ and $v^{\text{'}}=\cos x$

  • $y^{\text{'}}=\frac{-\sin x·\sin x-\cos x·\cos x}{{\left(\sin x\right)}^2}$

• Simplify using the identities ${\sin }^{2}x+\hspace{0.17em}{\cos }^{2}x=1$ and $\frac{1}{\sin x}=\csc x$

  • $y^{\text{'}}=\frac{-{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x}=\frac{-\left({\sin }^{2}x+{\cos }^{2}x\right)}{{\sin }^{2}x}=-\frac{1}{{\sin }^{2}x}=-{\csc }^{2}x$